It is now easy to see that \(\mathcal {S}\)  itself is an irreducible closed set:
If we cover \(\mathcal {S}\)  with two closed subsets then at least one of them
must contain infinitely many limit points. By \((\dagger )\)  that set
then is already all of \(\mathcal {S}\) . Of course, \(\mathcal {S}\)  is not the closure of a
singleton since there is no largest element, so we may conclude, as
Peter Johnstone did in [1]}, that \(\mathcal {S}\)  is not sober in
its Scott topology. In the terminology of the last section, it also
follows that \(\mathcal {S}\)  is not strongly complete. In preparation for our own counterexample, let us prove that \(\mathcal {S}\)  is the only closed subset which is not a point closure.
